Question 13.8: Adiabatic Mixing of Airstreams A stream of 5,000 cfm (2,359 ......
Adiabatic Mixing of Airstreams
A stream of 5,000 cfm (2,359 L/s) outdoor air at 40°F (4.4°C) dry-bulb temperature and 35°F (1.7°C) wet-bulb temperature is adiabatically mixed with 15,000 cfm (7,078 L/s) of recirculated air at 75°F (23.9°C) dry-bulb and 50% RH. Find the dry-bulb temperature and wet-bulb temperature of the resulting mixture.
Given: \dot{V}_{1} = 5000 cfm, T_{db1} = 40°F, T_{wb1} = 35°F, T_{db2} = 75°F, and \phi_{2} = 50%.
Figure: See Figure 13.9.
Assumptions: The atmospheric pressure is the standard atmosphere, 14.696 psia.
Find: T_{db3} and T_{wb3}.
Lookup values using psychrometric chart are as follows:
W_{1} = 0.00315 lb_{w}/lb_{a}, v_{1} = 122.65 ft^{3}/lb_{a},h_{1} = 13.2 Btu/lb_{a}
W_{2} = 0.00915 lb_{w}/lb_{a}, v_{2} = 122.65 ft^{3}/lb_{a},
h_{2} = 28.32 Btu/lb_{a}
Mass ow rate of stream 1 is
\dot{m}_{a1} = \frac{\dot{V}_{1}}{v_{1}} = \frac{5000 ft^{3}/min}{13.65 ft^{3}/lb_{a}} = 395.3 lb_{a}/minMass ow rate of stream 2 is
\dot{m}_{a1} = \frac{\dot{V}_{2}}{v_{2}} = \frac{15,000 ft^{3}/min}{13.68 ft^{3}/lb_{a}} = 1096.5 lb_{a}/minHumidity ratio of mixed air is
W_{3} = \frac{(\dot{m}_{a1} \cdot W_{1} + \dot{m}_{a2} \cdot W_{2})}{(\dot{m}_{a1} + \dot{m}_{a2})}= \frac{395.3 \times 0.0315 + 1096.5 \times 0.00915}{1491.8} = 0.00756 lb_{w}/lb_{a}
Similarly, the enthalpy of the mixture can be computed: h_{3} = 24.3 Btu/lb_{a}.
Finally, the dry-bulb temperature and the wetbulb temperature are found to be
T_{db3} = 66°F (18.9°C) and T_{wb3} = 56.5°F (13.6°C)